Answer:
The mass of FeSO4*7H2O is 1.023 grams
Explanation:
Step 1: Data given
Molar mass of FeSO4 * 7H2O = 278.01 g/mol
Mass of Fe2O3 = 0.294 grams
Mass of the sample = 3.055 grams
Step 2: The balanced equation
4 Fe + 3O2 → 2Fe2O3
Step 3: Calculate moles Fe2O3
Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3
Moles Fe2O3 = 0.294 grams / 159.59 g/mol
Moles Fe2O3 = 0.00184 moles
Step 3: Calculate moles Fe
In 2 moles Fe2O3 we have 4 moles Fe
For 0.00184 moles we'll have 2*0.00184 = 0.00368 moles
Step 4: The balanced equation
Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2
Step 5: Calculate moles FeSO4* 7H2O
For 1 mol Fe we have 1 mol FeSO4*7H2O
For 0.00368 moles Fe we have 0.00368 moles FeSO4*7H2O
Step 6: Calculate the mass of FeSO4*7H2O
Mass FeSO4*7H2O = moles * molar mass
Mass FeSO4*7H2O = 0.00368 moles * 278.01 g/mol
Mass FeSO4*7H2O = 1.023 grams
The mass of FeSO4*7H2O is 1.023 grams
